Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $a = \dfrac{7z - 7}{2} \times \dfrac{-7}{z^2 - z} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $a = \dfrac{ (7z - 7) \times -7 } { 2 \times (z^2 - z) } $ $ a = \dfrac {-7 \times 7(z - 1)} {2 \times z(z - 1)} $ $ a = \dfrac{-49(z - 1)}{2z(z - 1)} $ We can cancel the $z - 1$ so long as $z - 1 \neq 0$ Therefore $z \neq 1$ $a = \dfrac{-49 \cancel{(z - 1})}{2z \cancel{(z - 1)}} = -\dfrac{49}{2z} $